If 6x5 is exactly divisible by 9, then the least value of x is

Options

A. 1
B. 4
C. 7
D. 0

Answer

To solve the problem, we need to find the least value of \( x \) such that the number \( 6x5 \) is exactly divisible by \( 9 \). We will follow these steps:

### Step 1: Understand the divisibility rule for 9
A number is divisible by \( 9 \) if the sum of its digits is divisible by \( 9 \).

### Step 2: Identify the digits of the number
The number \( 6x5 \) has the digits \( 6 \), \( x \), and \( 5 \).

### Step 3: Calculate the sum of the digits
The sum of the digits can be expressed as:
\[ 6 + x + 5 = 11 + x \]

### Step 4: Set up the condition for divisibility
We want \( 11 + x \) to be divisible by \( 9 \). This means we need to find the smallest \( x \) such that:
\[ 11 + x \equiv 0 \mod 9 \]

### Step 5: Calculate \( 11 \mod 9 \)
Calculating \( 11 \mod 9 \):
\[ 11 \div 9 = 1 \quad \text{(remainder 2)} \]
So, \( 11 \equiv 2 \mod 9 \).

### Step 6: Set up the equation
We need:
\[ 2 + x \equiv 0 \mod 9 \]
This simplifies to:
\[ x \equiv -2 \mod 9 \]
or equivalently:
\[ x \equiv 7 \mod 9 \]

### Step 7: Find the least value of \( x \)
The smallest non-negative integer that satisfies \( x \equiv 7 \mod 9 \) is \( x = 7 \).

### Conclusion
Thus, the least value of \( x \) such that \( 6x5 \) is exactly divisible by \( 9 \) is:
\[ \boxed{7} \]

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